Sunday, December 30, 2007

NFL 2007: Pats & Parity

The regular season of the National Football League is now over and, far and away, the biggest story of the year is the New England Patriots' perfect 16-0 record.

As nearly all football fans would know, the last perfect regular season was the 1972 Miami Dolphins' 14-0 campaign; the Dolphins also swept their three post-season games to finish at 17-0. Two Chicago Bear teams from much earlier in NFL history, 1934 and 1942, also won all their regular-season games, but neither squad won the league title.

Before trying to figure out, post hoc, what the probability was of New England completing a 16-0 regular season, however, a brief look at the overall structure of wins and losses in the NFL is in order. If we were simply to flip coins to determine each team's record (i.e., heads = a win; tails = a loss), an infinite number of simulated 16-toss "seasons" would yield the following distribution.

As seen in the next graph, there were a few really good teams (the Colts, Cowboys, and Packers, each at 13-3, were just a cut below the Patriots), one dreadful team (the 1-15 Dolphins), and a lot of mediocre teams clustered around a record of 8-8, during this past season (click here for the final standings).

The actual collection of NFL team records this past season was thus fairly close -- disturbingly close in many fans' minds -- to what would be expected from just flipping coins!

OK, so what was the probability of the Patriots' perfect 16-0 regular season?

Using an approach I developed for my 35th anniversary retrospective on the Los Angeles Lakers' 33-game basketball winning streak during the 1971-72 season, I first attempted to estimate the Pats' chance of winning each of their games this past season, individually, based on the difficulty of the opposition in a particular contest and whether the game was played at home or on the road.

After looking at the final win-loss records of New England's 13 unique opponents (the Pats played their three intra-divisional rivals twice), I grouped the opponents into five levels of difficulty (click here for New England's schedule):

A (hardest opponents) -- The Patriots faced the aforementioned Colts and Cowboys, each of whom won 13 games.

B -- Teams that won 10 or 11 games, comprising San Diego, Cleveland, Pittsburgh, and the New York Giants, formed the second-toughest tier of opponents.

C -- Teams that won from 7 to 9 games, comprising Buffalo, Cincinnati, Washington, and Philadelphia, were deemed to be "mediocre" opponents.

D -- Teams that won 4 or 5 games, specifically the New York Jets and Baltimore, were considered "weak" opponents. The Ravens actually gave the Patriots one of their biggest scares, but that's neither here nor there, given my system of basing opponents' strength on objective win-loss records.

E -- The aforementioned Dolphins were in a class by themselves, providing the Patriots with their easiest opposition.

For each combination of opponent strength and home/away status, I came up with the following (assumed) probabilities of the Patriots' winning any given game (the guidelines below are similar, but not identical, to those I developed for the 1971-72 Lakers). To avoid any confusion, "home" refers to a game at New England.

E opponent at home ---> .95E opponent on the road ---> .90 D opponent at home ---> .85D opponent on the road ---> .80C opponent at home ---> .75C on road or B at home ---> .70B opponent on road ---> .65A opponent at home ---> .60A opponent on road ---> .55

The 16 individual game-specific Patriot-win probabilities were then multiplied together, according to what is known as the "multiplication rule." Multiplying these 16 probabilities together yielded .006; in other words, the chances of a perfect season like the Patriots' would be 6 out of 1,000 or roughly 1 in 167.

Again, this is just an estimate, based on some convenient assumptions, such as the outcomes of adjacent games being independent (i.e., winning one game does not affect the probability of winning the next game, beyond strength of opposition and game location). Were the distribution of win-loss records throughout the NFL in some future season to be substantially different from this year's (as graphed above), that would also affect the estimated probability, via the strength of opposition.

In any given season, just at a theoretical level, one or two teams (at most) might be expected to contend for 16-0. One must take into account the head-to-head aspect (especially within a conference); having teams play each other directly rules out the possibility of both going 16-0. To take the match-up in Super Bowl XIX (after the 1984 regular season) as an example, the NFC champion San Francisco 49ers had gone 15-1 in the regular season and the AFC champion Miami Dolphins had gone 14-2, so having a pair of teams threaten to go 16-0 in the same season is not totally farfetched.

Thus, if we held out the possibility of two teams per year possibly being able to go 16-0, then we would expect one team roughly every 85 years actually to do so (2 contenders per year X 85 years = 170, similar to the 1-in-167 figure I came up with, above). The 2007 NFL regular season was the 30th played under a 16-game format.

Next, we move on to see if New England can make it to -- and win -- the Super Bowl, and thus finish an unprecedented 19-0...